Home > Arrays, Iteration, Math > Generate maximum number from array of digits which is divisible by 2, 3 and 5

Generate maximum number from array of digits which is divisible by 2, 3 and 5

Whoa! Blogging 2 questions in one day! I guess I am actually in the mood to write tonight! Lets get started.

Let us take an example array, {9, 6, 3, 4}

So how should we go about solving for this example? Well, you don’t! Because if you notice, apart from 3, we are looking for divisibility by both 2 and 5. Which means that the final number should end in a “0”. And this array does not have a zero!

Moving on, here is the new array, {9, 6, 3, 4, 0}

Now what? Divisibility of 3 means that the sum of digits of the number should be divisible by 3. Note that you don’t have to use up all of the digits. So what do you do?

You can either go for an exponential approach by forming the power set of the array and shortlist the ones which have a sum divisible by 3 and then go on to see which of these has the most number of digits and even then if you have a tie, you’ll go for the one with the biggest digit.. and so on. But wait… I thought we were smart! Then why do all this?!

You can instead do the following –

1. Check if the array has atleast one 0 (zero). If not, we can’t form a number. Go back to bed.

2. Since you are still awake, go on to sort the array in descending order.

3. Now sum up the digits (and keep repeating Steps 3-6 till sum>=3). If your sum%3 == 0, we are done! This sequence is the biggest number. Sleep peacefully.

4. If not, there are 2 possibilities, sum%3 = 1 or 2 (let us call this remainder “R” which can have value 1 or 2)

5. We’ll first try to remove a number from this array which satisfies number%3==R starting from the smallest number in the array. As soon as you find such a number, remove it!

6. If not (life is such), go on to remove  2 numbers starting from the smallest number such that numbers%3==(3-R). Once you have removed 2 such numbers, you might have your desired number! Loop back to Step 3 and check the sum!

7. In the end you might be left with an array with only a 0. In that case your answer is either Aryabhata’s creation or “Cannot be formed” depending on the question’s requirement.

To quickly run through the example, {9, 6, 3, 4, 0}

In descending order, this is, {9, 6, 4, 3, 0} which sums to 22

22%3 = 1

Hence start trying to find a number%3=1 from the smallest,

3%3 = 1? No!

4%3 = 1? Yes!

Remove 4 and we are left with {9,6,3,0} or 9630 –> Which is the panacea for our metaphorical insomnia! Have fun! 🙂

Categories: Arrays, Iteration, Math Tags: ,
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